Using the equation $2 Al (s) + 3 H_{2} {SO}_{4} (a q) \to {Al}_{2} {({SO}_{4})}_{3} (a q) + 3 H_{2} (g)$ $2"Al"(s) + 3"H"_2"SO"_4(aq) -> "Al"_2("SO"_4)_3(aq) + 3"H"_2(g)$ . Starting with $15.0$ $15.0$ moles of $Al$ $"Al"$ , determine the theoretical yield of $H_{2}$ $"H"_2$ ?

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Using the equation $2 Al (s) + 3 H_{2} {SO}_{4} (a q) \to {Al}_{2} {({SO}_{4})}_{3} (a q) + 3 H_{2} (g)$ $2"Al"(s) + 3"H"_2"SO"_4(aq) -> "Al"_2("SO"_4)_3(aq) + 3"H"_2(g)$ . Starting with $15.0$ $15.0$ moles of $Al$ $"Al"$ , determine the theoretical yield of $H_{2}$ $"H"_2$ ?

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Answer 1 · 2018-02-26T03:53:54

${22.5 moles H}_{2}$ $"22.5 moles H"_2$

Explanation:

You first find the ratio between your unknown and your known, unknown on top and known on the bottom.

So with this question, your unknown is $H_{2}$ $"H"_2$ and your known is $Al$ $"Al"$ . Your ratio is $\frac{3}{2}$ $3/2$ .

Then, you take the number of moles of your known, $15$ $15$ moles of $Al$ $"Al"$ , and multiply it by that ratio and you get the number of moles of your unknown, which in this case is $22.5$ $22.5$ moles of $H_{2}$ $"H"_2$

$15 mol Al \cdot \frac{{3 mol H}_{2}}{2 mol Al} = {22.5 mol H}_{2}$ $"15 mol Al" * "3 mol H"_2/"2 mol Al" = "22.5 mol H"_2$

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Using the equation $2 Al (s) + 3 H_{2} {SO}_{4} (a q) \to {Al}_{2} {({SO}_{4})}_{3} (a q) + 3 H_{2} (g)$ $2"Al"(s) + 3"H"_2"SO"_4(aq) -> "Al"_2("SO"_4)_3(aq) + 3"H"_2(g)$ . Starting with $15.0$ $15.0$ moles of $Al$ $"Al"$ , determine the theoretical yield of $H_{2}$ $"H"_2$ ?

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Explanation:

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Using the equation 2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g)2"Al"(s) + 3"H"_2"SO"_4(aq) -> "Al"_2("SO"_4)_3(aq) + 3"H"_2(g). Starting with 15.015.0 moles of Al"Al", determine the theoretical yield of H2"H"_2 ?

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Using the equation $2 Al (s) + 3 H_{2} {SO}_{4} (a q) \to {Al}_{2} {({SO}_{4})}_{3} (a q) + 3 H_{2} (g)$ $2"Al"(s) + 3"H"_2"SO"_4(aq) -> "Al"_2("SO"_4)_3(aq) + 3"H"_2(g)$ . Starting with $15.0$ $15.0$ moles of $Al$ $"Al"$ , determine the theoretical yield of $H_{2}$ $"H"_2$ ?